Một vật dao động theo phương trình (t đo bằng giây). Tại thời điểm t1 li độ là cm và đang giảm. Tính li độ sau thời điểm t1 là 3 (s).
Một vật dao động theo phương trình (t đo bằng giây). Tại thời điểm t1 li độ là cm và đang giảm. Tính li độ sau thời điểm t1 là 3 (s).
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Tại thời điểm $t_1$, $x_1 = 2\sqrt{3}$ cm và vật đang giảm (v>0).
$2\sqrt{3} = 4cos(\frac{\pi t_1}{6})$ => $cos(\frac{\pi t_1}{6}) = \frac{\sqrt{3}}{2}$ => $\frac{\pi t_1}{6} = \frac{-\pi}{6}$ (do v>0) => $t_1 = -1$ (chọn nghiệm âm vì pha đang giảm)
Tại thời điểm $t_2 = t_1 + 3 = -1 + 3 = 2$ (s), ta có:
$x_2 = 4cos(\frac{\pi *2}{6}) = 4cos(\frac{\pi}{3}) = 4*\frac{1}{2} = 2$ cm.
Vì không có đáp án nào trùng với 2cm, ta kiểm tra lại điều kiện $t_1$.
$cos(\frac{\pi t_1}{6}) = \frac{\sqrt{3}}{2}$ => $\frac{\pi t_1}{6} = \frac{\pi}{6}$ => $t_1 = 1$
Tại thời điểm $t_2 = t_1 + 3 = 1 + 3 = 4$ (s), ta có:
$x_2 = 4cos(\frac{\pi *4}{6}) = 4cos(\frac{2\pi}{3}) = 4*(\frac{-1}{2}) = -2$ cm.
Vậy đáp án là C.
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